Integral problem: 1/((x^2)(sqrt(4-x^2)))

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In summary, the conversation discusses a problem involving integration using trigonometric substitution. The attempted solution involves substituting values and using a u-substitution, but results in incorrect steps. The correct solution involves using the trigonometric identity csc^2(θ) = 1/sin^2(θ) and simplifying the integral to get the final answer of -(√4-x^2/4x) + c.
  • #1
Palmira
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Homework Statement



∫ dx/ x2 (√4-x2)

Homework Equations


x=2sinθ
dx=2cosθdx
sin2 θ= (1 + cos (2θ))/2

The Attempt at a Solution


First attempt:
Plug in the trig sub and get:

∫2cosθdθ/4sin2θ2cosθ
The two cos θ cancel leaving

∫ dθ/ (4sin2θ)

Now I replace the sin2θ with (1-cos2θ)/2
and get

1/2 ∫dθ/(1-cos2θ)

I use u sub with the cos2θ
u=2θ
du/2=dθ

∫cosudu= sin2θ/2

I plug it back into the original equation

1/2 [ 2/θ-sin2θ]= 1/(θ-2sinθcosθ)

I retrieve the x from the original equation to get

4/ arcsin(x/2) -(x)(√(4-x2))

I may be doing something completely wrong, but I can't figure out where. Any hint on where to find the mistake? I keep practicing trig sub problems and I can not get any right. It might have to do with the derivation of the equations, but I am not sure.
 
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  • #2
Palmira said:

Homework Statement



∫ dx/ x2 (√4-x2)

Homework Equations


x=2sinθ
dx=2cosθdx
sin2 θ= (1 + cos (2θ))/2

The Attempt at a Solution


First attempt:
Plug in the trig sub and get:

∫2cosθdθ/4sin2θ2cosθ
The two cos θ cancel leaving

∫ dθ/ (4sin2θ)

Now I replace the sin2θ with (1-cos2θ)/2
and get

1/2 ∫dθ/(1-cos2θ)

I use u sub with the cos2θ
u=2θ
du/2=dθ

∫cosudu= sin2θ/2

I plug it back into the original equation

1/2 [ 2/θ-sin2θ]= 1/(θ-2sinθcosθ)

I retrieve the x from the original equation to get

4/ arcsin(x/2) -(x)(√(4-x2))

I may be doing something completely wrong, but I can't figure out where. Any hint on where to find the mistake? I keep practicing trig sub problems and I can not get any right. It might have to do with the derivation of the equations, but I am not sure.

When you say "I use u sub with the cos2θ" things start going badly wrong. I'm not even sure what you are doing after that. I would go back to when you have 1/(4sin^2(θ)). That's the same as csc^2(θ)/4. There's an easy integral for csc^2(θ).
 
  • #3
Well I do not feel very smart.

∫csc2xdx/ 4= -cot x/4 +c

go back to the triangle and it is -(√4-x2/ 4x) +c

Thank-you so much!
 
  • #4
Palmira said:
Well I do not feel very smart.

∫csc2xdx/ 4= -cot x/4 +c

go back to the triangle and it is -(√4-x2/ 4x) +c

Thank-you so much!

You're welcome and well done. You started ok. You just missed where to get off the substitution train.
 

Related to Integral problem: 1/((x^2)(sqrt(4-x^2)))

What is an integral problem?

An integral problem is a type of mathematical problem that involves finding the area under a curve, which is represented by a function. In other words, it is the process of finding the antiderivative of a given function.

Why is the integral of 1/((x^2)(sqrt(4-x^2))) difficult to solve?

The integral of 1/((x^2)(sqrt(4-x^2))) is difficult to solve because it is a type of improper integral, which means that it does not have a finite solution. This integral is also known as a trigonometric substitution, which requires advanced mathematical techniques to solve.

What is the significance of the denominator in the given integral problem?

The denominator, (x^2)(sqrt(4-x^2)), represents the area of a semicircle with radius 2. This is because the given function is in the form of 1/r^2, which is the formula for the area of a circle. The denominator also plays a crucial role in determining the convergence or divergence of the integral.

What are the possible techniques to solve the integral 1/((x^2)(sqrt(4-x^2)))?

There are several techniques that can be used to solve the given integral, such as trigonometric substitution, integration by parts, and partial fractions. Each technique may have its own advantages and limitations, so it is important to choose the most suitable one based on the given function.

What are the real-life applications of solving integral problems?

Integral problems have various real-life applications in fields such as physics, engineering, economics, and more. For example, finding the area under a velocity-time graph can help in determining the displacement of an object, which is crucial in understanding its motion. Integration is also used in calculating the work done by a force, the amount of fluid flowing through a pipe, and many other important concepts.

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